weird experiments schrödinger equation
DESCRIPTION
Weird experiments Schrödinger equation. Bohr model of an atom 1913. centrifugal is Latin for "center fleeing" It does not exist!. http://regentsprep.org/Regents/physics/phys06/bcentrif/centrif.htm. Bohr model of an atom 1913. Potential energy of the electron. - PowerPoint PPT PresentationTRANSCRIPT
Bohr model of an atom 19132
ecentrifugal
m vFr
http://regentsprep.org/Regents/physics/phys06/bcentrif/centrif.htm
Bohr model of an atom 1913
2e
centrifugalm vFr
2
Coulomb 20
eF4 r
Coulomb centrifugalF F
“Introduction to wave phenomena” by Akira Hirose and Karl Lonngren
Potential energy of the electron2
0
(J)4eU
r
22
204
em ver r
Bohr model of an atom 1913
22
0
1 (J)2 8e
em vr
Kinetic energy of the electron
22
204
em ver r
212 eU m vTotal energy of the electron
2
0
(J)8eE
r
2
04e
em e r
m vr
electron angular momentum
2 22 2 2
04e
em e r
m v rr
Bohr model of an atom 1913
electron angular momentum
Niels Bohr postulated that the momentum was quantized
( 1,2,3, )2ehm vr n n
22 2 110
2 5.3 10 (m)e
hr n n
m e
The radius is found to be
2
04e
em e r
m vr
h is Planck’s constant6.626068 × 10-34 m2 kg / s
2h
0
2
Bohr model of an atom 1913
http://csep10.phys.utk.edu/astr162/lect/light/bohr.html
The energy then becomes quantized
22 2 110
2 5.3 10 (m)e
hr n n
m e
4
2 2 20
2
18
1= -13.6 (eV)
en
m eE
h n
n
2
0
8eE
r
2
22 0
0 2
(J)8
e
ehn
m e
Photo electric effect - Einstein
http://regentsprep.org/Regents/physics/phys05/catomodel/bohr.htmHoudon
Energy of a photon E = h
2 1E - E h
Einstein’s explanation
19
34
14
2.9 10 J6.63 10 J sec
4.4 10 Hz
cW
vh
Bohr model of an atom 1913
What is the frequency of the light that will be emitted by an electron as it moves from the n = 2 down to n = 1?
2
1 -13.6 (eV)nE n
1 = -13.6 14
E h
Ionization implies n →
Experiment to understand the photo electric effect.
Experimental conclusions• The frequency must be greater than a “cut off
frequency” that changes with different metals.
• Kinetic energy of the emitted electrons depends upon the frequency of the incident light.
• Kinetic energy of the electrons is independent of the intensity of the incident light.
Sodium has a work function of W = 1.8 eV. Find the cutoff frequency.
cWh
144.4 10 Hz 19
34
1.8 1.6 10 J
6.63 10 J sec
cc
c
Å 6900
8
14
m3 10 sec4.4 10 Hz
76.9 10 m
A metal with a work function of 2.3 eV is illuminated with ultraviolet radiation = 3000 Ǻ. Calculate the energy of the
photo electrons that are emitted from the surface.
212 em v h W
hch
4.1 eV 34 8
7
6.63 10 3 10
3 10
196.63 10 J
21 4.1 2.32 em v 1.8 eV
Franck-Hertz experiment in mercury vapor. Electrons are accelerated and the current is monitored. 1914 (In 1887, Hertz noted that electrons would be emitted from a metal
that was illuminated with light.)
http://hyperphysics.phy-astr.gsu.edu/hbase/FrHz.html
2e 0
1 m v qV2
0e
e
2qVI n q Am
eI n qvA
Reflected wave is strong if n = 2d sin
dd sin
Davisson-Germer experiment – electrons incident on nickel 1925
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/davger2.html
Interpretation of the Davisson-Germer experiment
Energy of a photon E = h
gv k
2 Ehk
1 Ek
2
particleparticle
particle particle
1 mvE 2p (mv )
particlev
( 2 )k
Planckh2
de Broglie wavelengthwave energymomentumwave velocity
2mass velocity
velocity
hpc
h
de Broglie argued that there was a wavelength that could be written from
de Broglie hp
Interpretation of the Davisson-Germer experiment
particle
particle
E 1 Ep k
particlep kh 22
de Broglieparticle
hp
de Broglien 2d sin
Schrödinger equation
energy of particles2p U
2m
energy of photon hh ( 2 )2
2
deBroglie
h
U2m
2 2k U2m
2
deBroglie
2 h2
U2m
Schrödinger equation2 2k U2m
j( t kz )
0e
jt
jkz
22 2
2 jk kz
2 2k U2m
2 2
2j Ut 2m z
22j U
t 2m
Schrödinger equation2 2
2j Ut 2m z
22j U
t 2m
a2a1
( z,t ) * ( z,t )dzprobability
( z,t ) * ( z,t )dz
1probability of finding a state in aMax Born
2z a
Schrödinger equationa2a1
( z,t ) * ( z,t )dzprobability
( z,t ) * ( z,t )dz
j( t kz )
0e
elsewhere0
1 - 1 z 10
a1 a0 a2a1 a0 a2a1 a0 a2 a1 a0 a2
a0
+20-2
1
Schrödinger equation2 2
2( z,t ) ( z,t )j U ( z,t )t 2m z
( z,t ) Z( z )T( t ) 2 2
2dT( t ) d Z( z )j Z( z ) T( t ) UZ( z )T( t )dt 2m dz
2 2
21 dT( t ) 1 d Z( z )j U
T( t ) dt 2m Z( z ) dz
Schrödinger equationelectron in free space
2 2
21 dT( t ) 1 d Z( z )j U
T( t ) dt 2m Z( z ) dz
Ej t
0T( t ) T e
jkz jkzZ( z ) Ae Be2 2 2p kE U
2m 2m
2 2
21 dT( t ) 1 d Z( z )j
T( t ) dt 2m Z( z ) dz
Schrödinger equation
2 2
21 dT( t ) 1 d Z( z )j U( z )
T( t ) dt 2m Z( z ) dz
Ej t
0T( t ) T e Z( z ) Asin( kz ) Bcos( kz )
2 2 2p kE U2m 2m
Schrödinger equation
Ej t
0T( t ) T e
Z( z ) Asin( kz ) Bcos( kz )
Z(0 ) 0 B 0 nkL
Z( L ) 0
( z,t ) Z( z )T( t ) Ej t
0n zAT e sinL
L0
normalization ( z,t ) * ( z,t )dz 1
Schrödinger equation2
2j Ut 2m
2 2 22
2 2 2x y z
( x, y,z ) X ( x )Y( y )Z( z )
Schrödinger equation2
2j Ut 2m
( x, y,z ) X ( x )Y( y )Z( z )
Ej tyx z3
n yn x8 n z( x, y,z ) sin sin sin eL L LL
2 22 22yx znn nE U
8mL L L L
Schrödinger equationEj tyx z
3
n yn x8 n z( x, y,z ) sin sin sin eL L LL
Schrödinger equation
22j U
t 2m
2
sin
sin sin
22
22 2 2 2
r1 1 1r
rr r r
( r , , ) R( r ) ( ) ( )
Schrödinger equation
Schrödinger equation
element n l m s
Hydrogen 1 0 0 +1/2 or -1/2
Helium 1 0 0 +1/2 & -1/2
Beryllium 2 0 0 +1/2 & -1/2Lithium 2 0 0 +1/2 or -1/2
Heisenberg uncertainty principle
http://www.aip.org/history/heisenberg/
( position ) ( momentum ) hx p h
( energy ) ( time ) hE t h
2 2m mv v v h2 2
mv v h vh
m v h
x(m v ) m v h